Resolução eletromagnetismo - hayt - 6 ed

CHAPTER 1 1.1. Given the vectors M = −10ax + 4ay − 8az and N = 8ax + 7ay − 2az , find: a) a unit vector in the direction of −M + 2N. −M + 2N = 10ax − 4ay + 8az + 16ax + 14ay − 4az = (26, 10, 4) Thus a= b) the magnitude of 5ax + N − 3M: (5, 0, 0) + (8, 7, −2) − (−30, 12, −24) = (43, −5, 22), and |(43, −5, 22)| = 48.6. c) |M||2N|(M + N): |(−10, 4, −8)||(16, 14, −4)|(−2, 11, −10) = (13.4)(21.6)(−2, 11, −10) = (−580.5, 3193, −2902) 1.2. Given three points, A(4, 3, 2), B(−2, 0, 5), and C(7, −2, 1): a) Specify the vector A extending from the origin to the point A. A = (4, 3, 2) = 4ax + 3ay + 2az b) Give a unit vector extending from the origin to the midpoint of line AB. The vector from the origin to the midpoint is given by M = (1/2)(A + B) = (1/2)(4 − 2, 3 + 0, 2 + 5) = (1, 1.5, 3.5) The unit vector will be m= (1, 1.5, 3.5) = (0.25, 0.38, 0.89) |(1, 1.5, 3.5)| (26, 10, 4) = (0.92, 0.36, 0.14) |(26, 10, 4)|

c) Calculate the length of the perimeter of triangle ABC: Begin with AB = (−6, −3, 3), BC = (9, −2, −4), CA = (3, −5, −1). Then |AB| + |BC| + |CA| = 7.35 + 10.05 + 5.91 = 23.32 1.3. The vector from the origin to the point A is given as (6, −2, −4), and the unit vector directed from the origin toward point B is (2, −2, 1)/3. If points A and B are ten units apart, find the coordinates of point B. With A = (6, −2, −4) and B = 1 B(2, −2, 1), we use the fact that |B − A| = 10, or 3 |(6 − 2 B)ax − (2 − 2 B)ay − (4 + 1 B)az | = 10 3 3 3 Expanding, obtain 36 − 8B + 4 B 2 + 4 − 8 B + 4 B 2 + 16 + 8 B + 1 B 2 = 100 9 3 9 3 9 or B 2 − 8B − 44 = 0. Thus B = B=
√ 8± 64−176 2

= 11.75 (taking positive option) and so

2 1 2 (11.75)ax − (11.75)ay + (11.75)az = 7.83ax − 7.83ay + 3.92az 3 3 3 1

1.4. given points A(8, −5, 4) and B(−2, 3, 2), find: a) the distance from A to B. |B − A| = |(−10, 8, −2)| = 12.96 b) a unit vector directed from A towards B. This is found through aAB = B−A = (−0.77, 0.62, −0.15) |B − A|

c) a unit vector directed from the origin to the midpoint of